Question

Find the locus of the middle points of chords of an ellipse whose length is constant $(=2c)$.

Answer 1

Suppose the coordinates of the middle point of the chord $PQ$ of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ be $(\alpha ,\beta )$

As length of $PQ$ is $2c$ So the coordinates of $P$ and $Q$ may be taken as $(\alpha +c\cos \theta ,\beta +c\sin \theta )$ and $(\alpha -c\cos \theta ,\beta -c\sin \theta )$

$P,Q$ points lie on ellipse

$\frac{{(\alpha +c\cos \theta )}^2}{a^2}+\frac{{(\beta +c\sin \theta )}^2}{b^2}=1⟶1$

$\frac{{(\alpha -c\cos \theta )}^2}{a^2}+\frac{{(\beta -c\sin \theta )}^2}{b^2}=1⟶2$

$1+2⟹\frac{2}{a^2}({\alpha }^2+c^2{\cos }^2\theta )+\frac{2}{b^2}({\beta }^2+c^2{\sin }^2\theta )=2$

$⟹b^2{\alpha }^2+a^2{\beta }^2-a^2{b}^2+c^2(a^2{\sin }^2\theta +b^2{\cos }^2\theta )=0⟶3$

$1-2⟹\frac{4\alpha c\cos \theta }{a^2}+\frac{4\beta c\sin \theta }{b^2}=0$

$⟹\frac{\sin \theta }{{\alpha b}^2}=\frac{\cos \theta }{{-\beta a}^2}=\frac{1}{\sqrt{{\alpha }^2{b}^4+{\beta }^2{a}^4}}$

$\sin \theta =\frac{{\alpha b}^2}{\sqrt{{\alpha }^2{b}^4+{\beta }^2{a}^4}},\cos \theta =\frac{{-\beta a}^2}{\sqrt{{\alpha }^2{b}^4+{\beta }^2{a}^4}}$

From $3,$ $b^2{\alpha }^2+a^2{\beta }^2-a^2{b}^2+c^2\left(\frac{{{\alpha }^2{b}^{4}a}^2-{{\beta }^2{a}^{4}b}^2}{{\left(\sqrt{{\alpha }^2{b}^4+{\beta }^2{a}^4}\right)}^2}\right)=0$

$⟹({\alpha }^2{b}^4+{\beta }^2{a}^4)({\alpha }^2{b}^2+{\beta }^2{a}^2-a^2{b}^2)+c^2{a}^2{b}^2({\alpha }^2{b}^2+{\beta }^2{a}^2)=0$

By generalizing, the locus of $(\alpha ,\beta )$ is

$(x^2{b}^4+y^2{a}^4)(x^2{b}^2+y^2{a}^2-a^2{b}^2)+c^2{a}^2{b}^2(x^2{b}^2+y^2{a}^2)=0$

which is an ellipse